#include "singularvaluedecomposition.h"
#include "matrix.h"

#include <cmath>

using namespace std;

using namespace CPAMA;

SingularValueDecomposition::SingularValueDecomposition(Matrix Arg) {
    // Derived from LINPACK code.
    // Initialize.
    double** A = Arg.getArrayCopy();
    m = Arg.getRowDimension();
    n = Arg.getColumnDimension();

    /* Apparently the failing cases are only a proper subset of (m<n),
   so let's not throw error.  Correct fix to come later?
    if (m<n) {
    throw new IllegalArgumentException("Jama SVD only works for m >= n"); }
    */
    int nu = fmin(m,n);
    slen = fmin(m+1,n);
    s = new double [slen];
    U = new double*[m];
    for (int i = 0; i < m; i++) {
        U[i] = new double[nu];
    }
    V = new double*[n];
    for (int i = 0; i < n; i++) {
        V[i] = new double[n];
    }
    double* e = new double [n];
    double* work = new double [m];
    bool wantu = true;
    bool wantv = true;

    // Reduce A to bidiagonal form, storing the diagonal elements
    // in s and the super-diagonal elements in e.

    int nct = fmin(m-1,n);
    int nrt = fmax(0,fmin(n-2,m));
    for (int k = 0; k < fmax(nct,nrt); k++) {
       if (k < nct) {

          // Compute the transformation for the k-th column and
          // place the k-th diagonal in s[k].
          // Compute 2-norm of k-th column without under/overflow.
          s[k] = 0;
          for (int i = k; i < m; i++) {
             s[k] = hypot(s[k],A[i][k]);
          }
          if (s[k] != 0.0) {
             if (A[k][k] < 0.0) {
                s[k] = -s[k];
             }
             for (int i = k; i < m; i++) {
                A[i][k] /= s[k];
             }
             A[k][k] += 1.0;
          }
          s[k] = -s[k];
       }
       for (int j = k+1; j < n; j++) {
          if ((k < nct) & (s[k] != 0.0))  {

          // Apply the transformation.

             double t = 0;
             for (int i = k; i < m; i++) {
                t += A[i][k]*A[i][j];
             }
             t = -t/A[k][k];
             for (int i = k; i < m; i++) {
                A[i][j] += t*A[i][k];
             }
          }

          // Place the k-th row of A into e for the
          // subsequent calculation of the row transformation.

          e[j] = A[k][j];
       }
       if (wantu & (k < nct)) {

          // Place the transformation in U for subsequent back
          // multiplication.

          for (int i = k; i < m; i++) {
             U[i][k] = A[i][k];
          }
       }
       if (k < nrt) {

          // Compute the k-th row transformation and place the
          // k-th super-diagonal in e[k].
          // Compute 2-norm without under/overflow.
          e[k] = 0;
          for (int i = k+1; i < n; i++) {
             e[k] = hypot(e[k],e[i]);
          }
          if (e[k] != 0.0) {
             if (e[k+1] < 0.0) {
                e[k] = -e[k];
             }
             for (int i = k+1; i < n; i++) {
                e[i] /= e[k];
             }
             e[k+1] += 1.0;
          }
          e[k] = -e[k];
          if ((k+1 < m) & (e[k] != 0.0)) {

          // Apply the transformation.

             for (int i = k+1; i < m; i++) {
                work[i] = 0.0;
             }
             for (int j = k+1; j < n; j++) {
                for (int i = k+1; i < m; i++) {
                   work[i] += e[j]*A[i][j];
                }
             }
             for (int j = k+1; j < n; j++) {
                double t = -e[j]/e[k+1];
                for (int i = k+1; i < m; i++) {
                   A[i][j] += t*work[i];
                }
             }
          }
          if (wantv) {

          // Place the transformation in V for subsequent
          // back multiplication.

             for (int i = k+1; i < n; i++) {
                V[i][k] = e[i];
             }
          }
       }
    }

    // Set up the final bidiagonal matrix or order p.

    int p = fmin(n,m+1);
    if (nct < n) {
       s[nct] = A[nct][nct];
    }
    if (m < p) {
       s[p-1] = 0.0;
    }
    if (nrt+1 < p) {
       e[nrt] = A[nrt][p-1];
    }
    e[p-1] = 0.0;

    // If required, generate U.

    if (wantu) {
       for (int j = nct; j < nu; j++) {
          for (int i = 0; i < m; i++) {
             U[i][j] = 0.0;
          }
          U[j][j] = 1.0;
       }
       for (int k = nct-1; k >= 0; k--) {
          if (s[k] != 0.0) {
             for (int j = k+1; j < nu; j++) {
                double t = 0;
                for (int i = k; i < m; i++) {
                   t += U[i][k]*U[i][j];
                }
                t = -t/U[k][k];
                for (int i = k; i < m; i++) {
                   U[i][j] += t*U[i][k];
                }
             }
             for (int i = k; i < m; i++ ) {
                U[i][k] = -U[i][k];
             }
             U[k][k] = 1.0 + U[k][k];
             for (int i = 0; i < k-1; i++) {
                U[i][k] = 0.0;
             }
          } else {
             for (int i = 0; i < m; i++) {
                U[i][k] = 0.0;
             }
             U[k][k] = 1.0;
          }
       }
    }

    // If required, generate V.

    if (wantv) {
       for (int k = n-1; k >= 0; k--) {
          if ((k < nrt) & (e[k] != 0.0)) {
             for (int j = k+1; j < nu; j++) {
                double t = 0;
                for (int i = k+1; i < n; i++) {
                   t += V[i][k]*V[i][j];
                }
                t = -t/V[k+1][k];
                for (int i = k+1; i < n; i++) {
                   V[i][j] += t*V[i][k];
                }
             }
          }
          for (int i = 0; i < n; i++) {
             V[i][k] = 0.0;
          }
          V[k][k] = 1.0;
       }
    }

    // Main iteration loop for the singular values.

    int pp = p-1;
    int iter = 0;
    double eps = pow(2.0,-52.0);
    double tiny = pow(2.0,-966.0);
    while (p > 0) {
       int k,kase;

       // Here is where a test for too many iterations would go.

       // This section of the program inspects for
       // negligible elements in the s and e arrays.  On
       // completion the variables kase and k are set as follows.

       // kase = 1     if s(p) and e[k-1] are negligible and k<p
       // kase = 2     if s(k) is negligible and k<p
       // kase = 3     if e[k-1] is negligible, k<p, and
       //              s(k), ..., s(p) are not negligible (qr step).
       // kase = 4     if e(p-1) is negligible (convergence).

       for (k = p-2; k >= -1; k--) {
          if (k == -1) {
             break;
          }
          if (abs(e[k]) <=
                tiny + eps*(abs(s[k]) + abs(s[k+1]))) {
             e[k] = 0.0;
             break;
          }
       }
       if (k == p-2) {
          kase = 4;
       } else {
          int ks;
          for (ks = p-1; ks >= k; ks--) {
             if (ks == k) {
                break;
             }
             double t = (ks != p ? abs(e[ks]) : 0.) +
                        (ks != k+1 ? abs(e[ks-1]) : 0.);
             if (abs(s[ks]) <= tiny + eps*t)  {
                s[ks] = 0.0;
                break;
             }
          }
          if (ks == k) {
             kase = 3;
          } else if (ks == p-1) {
             kase = 1;
          } else {
             kase = 2;
             k = ks;
          }
       }
       k++;

       // Perform the task indicated by kase.

       switch (kase) {

          // Deflate negligible s(p).

          case 1: {
             double f = e[p-2];
             e[p-2] = 0.0;
             for (int j = p-2; j >= k; j--) {
                double t = hypot(s[j],f);
                double cs = s[j]/t;
                double sn = f/t;
                s[j] = t;
                if (j != k) {
                   f = -sn*e[j-1];
                   e[j-1] = cs*e[j-1];
                }
                if (wantv) {
                   for (int i = 0; i < n; i++) {
                      t = cs*V[i][j] + sn*V[i][p-1];
                      V[i][p-1] = -sn*V[i][j] + cs*V[i][p-1];
                      V[i][j] = t;
                   }
                }
             }
          }
          break;

          // Split at negligible s(k).

          case 2: {
             double f = e[k-1];
             e[k-1] = 0.0;
             for (int j = k; j < p; j++) {
                double t = hypot(s[j],f);
                double cs = s[j]/t;
                double sn = f/t;
                s[j] = t;
                f = -sn*e[j];
                e[j] = cs*e[j];
                if (wantu) {
                   for (int i = 0; i < m; i++) {
                      t = cs*U[i][j] + sn*U[i][k-1];
                      U[i][k-1] = -sn*U[i][j] + cs*U[i][k-1];
                      U[i][j] = t;
                   }
                }
             }
          }
          break;

          // Perform one qr step.

          case 3: {

             // Calculate the shift.

             double scale = fmax(fmax(fmax(fmax(
                     abs(s[p-1]),abs(s[p-2])),abs(e[p-2])),
                     abs(s[k])),abs(e[k]));
             double sp = s[p-1]/scale;
             double spm1 = s[p-2]/scale;
             double epm1 = e[p-2]/scale;
             double sk = s[k]/scale;
             double ek = e[k]/scale;
             double b = ((spm1 + sp)*(spm1 - sp) + epm1*epm1)/2.0;
             double c = (sp*epm1)*(sp*epm1);
             double shift = 0.0;
             if ((b != 0.0) | (c != 0.0)) {
                shift = sqrt(b*b + c);
                if (b < 0.0) {
                   shift = -shift;
                }
                shift = c/(b + shift);
             }
             double f = (sk + sp)*(sk - sp) + shift;
             double g = sk*ek;

             // Chase zeros.

             for (int j = k; j < p-1; j++) {
                double t = hypot(f,g);
                double cs = f/t;
                double sn = g/t;
                if (j != k) {
                   e[j-1] = t;
                }
                f = cs*s[j] + sn*e[j];
                e[j] = cs*e[j] - sn*s[j];
                g = sn*s[j+1];
                s[j+1] = cs*s[j+1];
                if (wantv) {
                   for (int i = 0; i < n; i++) {
                      t = cs*V[i][j] + sn*V[i][j+1];
                      V[i][j+1] = -sn*V[i][j] + cs*V[i][j+1];
                      V[i][j] = t;
                   }
                }
                t = hypot(f,g);
                cs = f/t;
                sn = g/t;
                s[j] = t;
                f = cs*e[j] + sn*s[j+1];
                s[j+1] = -sn*e[j] + cs*s[j+1];
                g = sn*e[j+1];
                e[j+1] = cs*e[j+1];
                if (wantu && (j < m-1)) {
                   for (int i = 0; i < m; i++) {
                      t = cs*U[i][j] + sn*U[i][j+1];
                      U[i][j+1] = -sn*U[i][j] + cs*U[i][j+1];
                      U[i][j] = t;
                   }
                }
             }
             e[p-2] = f;
             iter = iter + 1;
          }
          break;

          // Convergence.

          case 4: {

             // Make the singular values positive.

             if (s[k] <= 0.0) {
                s[k] = (s[k] < 0.0 ? -s[k] : 0.0);
                if (wantv) {
                   for (int i = 0; i <= pp; i++) {
                      V[i][k] = -V[i][k];
                   }
                }
             }

             // Order the singular values.

             while (k < pp) {
                if (s[k] >= s[k+1]) {
                   break;
                }
                double t = s[k];
                s[k] = s[k+1];
                s[k+1] = t;
                if (wantv && (k < n-1)) {
                   for (int i = 0; i < n; i++) {
                      t = V[i][k+1]; V[i][k+1] = V[i][k]; V[i][k] = t;
                   }
                }
                if (wantu && (k < m-1)) {
                   for (int i = 0; i < m; i++) {
                      t = U[i][k+1]; U[i][k+1] = U[i][k]; U[i][k] = t;
                   }
                }
                k++;
             }
             iter = 0;
             p--;
          }
          break;
       }
    }

    delete[] e;
    delete[] work;

}

SingularValueDecomposition::~SingularValueDecomposition() {
    delete[] U;
    delete[] V;
    delete[] s;
}

Matrix SingularValueDecomposition::getU () {
   return Matrix(U,m,fmin(m+1,n));
}

Matrix SingularValueDecomposition::getV () {
   return Matrix(V,n,n);
}

double* SingularValueDecomposition::getSingularValues () {
   return s;
}

Matrix SingularValueDecomposition::getS () {
   Matrix X = Matrix(n,n);
   double** S = X.getArray();
   for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
         S[i][j] = 0.0;
      }
      S[i][i] = this->s[i];
   }
   return X;
}

double SingularValueDecomposition::norm2 () {
   return s[0];
}

double SingularValueDecomposition::cond () {
   return s[0]/s[(int)fmin(m,n)-1];
}

int SingularValueDecomposition::rank () {
   double eps = pow(2.0,-52.0);
   double tol = fmax(m,n)*s[0]*eps;
   int r = 0;
   for (int i = 0; i < slen; i++) {
      if (s[i] > tol) {
         r++;
      }
   }
   return r;
}
